A status page logs each hour as 1 (the service was up) or 0 (it was down). An engineer can mark at most k down hours as patched, turning them into 1s.
Write longestUptime(hours, k) that returns the length of the longest run of neighboring hours that are all 1 after patching at most k of the 0s.
Input: hours = [1,1,0,0,1,1,1,0,1], k = 1 Output: 5
Patch the 0 at index 7 to get 1, 1, 1, 1, 1 from index 4 to 8.
Input: hours = [0,0,0], k = 0 Output: 0
No patches and no up hours.
0 <= hours.length <= 100000hours[i] is 0 or 1.0 <= k <= hours.lengthRun tries your code on the examples. Submit runs every test, including hidden edge cases, and solves the problem when they all pass.
Ctrl + Enter runs, Ctrl + Shift + Enter submits (⌘ on a Mac).
I'm here to help with #137 Longest Uptime With k Patches. I give hints, never the solution, so you get to solve it yourself. Ask for a hint, select code to ask about it, or press Ask Tuto why on a failing test.
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